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In the case of water (H20) we have two molecules of hydrogen and one of oxygen converted into two molecules of water.

There are four possible cases to be considered depending on the direction of rotation of the several molecules.

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Hence, if 4€, be the atomic energy lost during collision,

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From (A) and (B) we find, by eliminating w", and leaving the sign of in to be determined by the resulting equation,

Case I.

ma + 16m + 192 = 0.

Case II. / The roots of this equation are imaginary, and, therefore, these cases impossible.

Case III.

m2 + 16m + 96 = 0. Case IV. , "

The roots of this equation also are imaginary, and these cases impossible.

It is evident, from the foregoing, that it is impossible to satisfy equations (A) and (B) so long as the value of m is the same in both.

This impossibility of hydrogen and oxygen combining together of themselves to form water is entirely in accordance with chemical facts. If a mixture of two volumes of hydrogen and one volume of oxygen be made, they will not combine unless aided from without by a lighted match or an electric spark; when this aid is given, water is at once formed with an explosion, which corresponds to a “catastrophe" in a planetary system.

We must now modify equations (A) and (B) so as to suit the condition of aid from without, and give a real value of m.

Equation (B) which depends solely on geometrical configuration does not admit of modification ; but equation (A) which represents the conservation of areas can be easily modified ; let us suppose that the effect of the lighted match or electric spark is to add a quantity a to the right-hand side of the equation of areas. We now have Case I.

8

= 1 t-ta
Case II.
Case III.

= t-ta',

Case IV. where a and a' are the areas necessary to be added; from which we see that

a' = 1 + a. Cases I. and II. are therefore to be preferred to Cases III. and IV., because they require a less amount of external aid.

We now have

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I have already shown that in the case of hydroxyl the most pro

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Now, since H,0 > HO, we have

w'2 > 1.2564,
m = - 10,

W2 = + 1.3600,
Az = 1.3090,

Ez = + 0.0984.
m = -11,

w = 1.2066,
Az = 1.2412,

E2 = -0.0217.
The last of these must be rejected, for ez must be positive.
The first fulfils the essential conditions

W?>w?, az < az.
Therefore select m=- 10.
As this differs from my=- 7

in other words, the attraction of oxygen upon hydrogen depends upon its having one or two satellites of hydrogen to deal with. In fact, we find

r = 25829,

M'' = 5625, where u' is the coefficient of attraction of oxygen upon hydrogen in hydroxyl, and fe" the coefficient of attraction of oxygen upon hydrogen in water.

2. Sulphuretted Hydrogen. Sulphur, selenium, and tellurium, behave towards hydrogen in the same manner as oxygen ; and chemists recognize HS analogous to hydroxyl, and H.S (sulphuretted hydrogen) analogous to water.

Like hydroxyl, HS is not known to exist as a separate compound, but it may exist in combination in higher compounds. Introducing B = 32 in the general equation (7) bis, we find

(66€1 + 1) m2 – 2048m - 31712 = 0.

If € = 0, we find m2 =- 15, the nearest whole number to the negative root.

From this we find

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-1.0524,

w"2 = 1·1076, a= 07687. It is easy to show, as in the case of oxygen, that no dyad atom can enter into combination with two hydrogen atoms without external aid ; for, if ß be the atomic weight of any dyad, we have Conservation of areas

Case I. w" = 1 +
Case II.

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Both equations give imaginary values of m for real values of ß. We must therefore suppose, as in the case of water, that an addition a is made to the right-hand side of the equation of conservation of areas. This gives us,

12m + BV Areas, . .

E + a. | 2mi

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